If the diameter of the circle 2(x^2 + y^2) + | vedclass

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(D) The given equation of the circle is $2x^2 + 2y^2 + 3x + 4y - 1 = 0$.Dividing by $2$,we get $x^2 + y^2 + \frac{3}{2}x + 2y - \frac{1}{2} = 0$.The c

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$1/2$

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(D) The given equation of the circle is $2x^2 + 2y^2 + 3x + 4y - 1 = 0$.Dividing by $2$,we get $x^2 + y^2 + \frac{3}{2}x + 2y - \frac{1}{2} = 0$.The center of the circle $(h, k')$ is given by $(-\frac{g}{2}, -\frac{f}{2})$,where $2g = \frac{3}{2}$ and $2f = 2$.Thus,the center is $(-\frac{3}{4}, -1)$.Since the diameter $y = 2x + k$ passes through the center of the circle,we substitute the center coordinates into the equation:$-1 = 2(-\frac{3}{4}) + k$$-1 = -\frac{3}{2} + k$$k = -1 + \frac{3}{2} = \frac{1}{2}$.

If the diameter of the circle $2(x^2 + y^2) + 3x + 4y - 1 = 0$ is $y = 2x + k$,then $k = \dots$

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